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A kac moody verma module is not integrable in general
Statement refuted
False claim: a Kac–Moody Verma module of dominant integral highest weight is integrable.
For every finite GCM with at least one simple index and every , its Verma highest vector satisfies for every and every simple index . Thus no such Verma module is integrable, even at dominant integral . The concrete witness is , .
Facts & Assumptions
Given: A finite GCM with at least one simple index and its Verma module.
Verma modules are nonzero highest-weight modules with highest vector (Universal property and pbw character of kac moody verma modules).
Integrability requires local nilpotence of every (Integrable kac moody module).
The induced tensor construction identifies with , sending to (Kac moody verma module).
Ordered monomials form a basis for the enveloping algebra, with homogeneous bases available without AC (PBW for countably presented Kac Moody Lie algebras).
The simple negative root space is the nonzero line (Kac moody root spaces are finite dimensional).
Counterexample
Fix a simple index . By F5 its simple negative root space is one dimensional with specified nonzero vector . Use the homogeneous basis in F4, replacing its vector at that degree by if needed; this is a single nonzero rescaling. For every , the ordered monomial consisting of copies of is a PBW basis monomial by F4, including the empty monomial at . Hence these powers are nonzero and independent in . F3 identifies them with , so none vanishes.
By 1.1, the single nonzero vector from F1 is killed by no power of . This fails the local nilpotence requirement F2 and proves nonintegrability. For the explicit witness , , the label zero is dominant integral, while are all nonzero independent vectors. Thus the extra dominant-weight hypothesis does not repair the false claim. This includes the zeroth and first powers; there is no terminal power. A GCM with no simple indices is excluded, since then there is no lowering operator to witness failure. The basis construction and one specified rescaling require no AC.
Depends on
Used by
- Local nilpotence of only the ei does not imply integrability Counterexample
Dependency tree · two levels
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Sources
- Kleshchev, Lectures on Infinite Dimensional Lie Algebras (standard reference, not scraped)
- Perrin, Introduction to Kac-Moody Groups and Lie Algebras (standard reference, not scraped)